PHP Error : Notice: Undefined variable

Question

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PHP: “Notice: Undefined variable” and “Notice: Undefined index”

I want when I choose some value in a Select box to display a string for a number of times for example if I choosed the value 2 it will display a string twice.

this is the code I tried :

<!DOCTYPE HTML>
<html>
<head>
<script>
function checkIt()
{
    var getNum = document.getElementById("numb").value;

    //this_file.php ... im specifying for just. you specify this full code in any of file and specify the whole url path
    location.href="this_file.php?countIt="+getNum;
}
</script>
    <style type="text/css">
    .centrer
    {
        position: absolute;
        left: 50%;
        top: 50%;
        font-size:24px;
    }
    </style>

</head>
<body>
    <select name="nombres" id="numb" onchange="checkIt();">
        <option>1</option>
        <option>2</option>
        <option>3</option>
    </select>

<?php
if($_REQUEST["countIt"])
{
    $displayTimes = $_REQUEST["countIt"]; 
}
?>
   <div class="centrer">
    <?php
        $s = "Hello World! <br/>";
        for($i=0; $i<$displayTimes; $i++)
            echo $s;
    ?>
    </div>
</body>
</html>

but I have a problem :

Notice: Undefined variable: displayTimes in C:\Program Files\EasyPHP-12.1\www\site\Untitled-1.php on line 43

notice that the line 43 is : for($i=0; $i<$displayTimes; $i++)

and also another problem that when I select the "1" Value it does anything, and also when I select other value the selected value in the select box still the value "1" and I have another question isn't there any other way to do this without using JS only by using PHP ?

Well `$displayTimes` is undefined as php states – zerkms Oct 18 '12 at 23:27 — zerkms, Oct 18 '12 at 23:27

score 6 · Accepted Answer · edited Oct 18 '12 at 23:34

6

$displayTimes would only be declared if the condition was met, it was caused by the following :

if($_REQUEST["countIt"]) <------------------- Condition 
{
    $displayTimes = $_REQUEST["countIt"];
}

Replace the above with :

$displayTimes = isset($_REQUEST["countIt"]) ? $_REQUEST["countIt"] : 0;

edited Oct 18 '12 at 23:34

Musa

96,336
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answered Oct 18 '12 at 23:28

Baba

94,024
28
166
217

2

And the next bug OP will find is that his/her file isn't named **this_file.php** :) – cbayram Oct 18 '12 at 23:36
@Baba what is the role of isset ? – Renaud is Not Bill Gates Oct 18 '12 at 23:57
[isset — Determine if a variable is set and is not NULL](http://php.net/manual/en/function.isset.php) – Baba Oct 19 '12 at 00:00
@Baba thanks for your help but I still have the problem that when I select the "1" Value it does anything, and also when I select other value the selected value in the select box still the value "1" – Renaud is Not Bill Gates Oct 19 '12 at 00:07
Just add `` as the first option .... – Baba Oct 19 '12 at 00:10
@Baba one more only question : Is there any other way without using JavaScript only by using php ? – Renaud is Not Bill Gates Oct 19 '12 at 00:23
PHP runs on server while JavaScript Runs on Browsers or Nodejs server .. if you want to use only PHP you would have to separate the pages and post to the other page with submit – Baba Oct 19 '12 at 00:26
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/18265/discussion-between-su-sha-and-baba) – Renaud is Not Bill Gates Oct 19 '12 at 01:07

score 1 · Answer 2 · answered Oct 18 '12 at 23:29

The variable $displayTimes must be set in order to use it. If $_REQUEST["countIt"] isn't set, then $displayTimes isn't set.

Change it to the following:

$displayTimes = isset($_REQUEST['countIt']) ? $_REQUEST['countIt'] : 0;

This way, if $_REQUEST['countIt'] isn't set it won't be executed.

Marian Zburlea · Answer 3 · 2012-10-19T00:12:06.913

1

This is a more clean code:

<?php
if(isset($_GET['countIt'])) {
  print '<div class="centrer">';
  $s = "Hello World! <br/>";
  for($i=0; $i<$_GET['countIt']; ++$i) echo $s;
  print '</div>';
}
?>

And check this JS code: http://jsfiddle.net/xPuEN/1/

edited Oct 19 '12 at 00:12

answered Oct 18 '12 at 23:35

Marian Zburlea

9,177
4
31
37

Thanks, but I sill have the problem that when I select the "1" Value it does anything, and also when I select other value the selected value in the select box still the value "1" – Renaud is Not Bill Gates Oct 19 '12 at 00:07

PHP Error : Notice: Undefined variable

3 Answers3